Bourbaki

Exercise 1, § 1

General Topology, VIII

Let f(z)=zn+a1zn−1+⋯+an−1z+anf(z) = z^n + a_1 z^{n-1} + \cdots + a_{n-1} z + a_n be a polynomial of degree nn with complex coefficients and put f(z)=∏i=1n(z−zi)f(z) = \prod_{i=1}^n (z - z_i). Let r0=sup⁡i∣zi∣r_0 = \sup_i |z_i|.

a) Show that if the real number r>0r > 0 is such that

rn≤∣a1∣rn−1+∣a2∣rn−2+⋯+∣an−1∣r+∣an∣, r^n \leq |a_1|r^{n-1} + |a_2|r^{n-2} + \cdots + |a_{n-1}|r + |a_n|,
then r0≤rr_0 \leq r, and deduce that
r0≤sup⁡(1,∑k=1n∣ak∣). r_0 \leq \sup \left( 1, \sum_{k=1}^n |a_k| \right).

b) Let (λi)1≤i≤n(\lambda_i)_{1 \leq i \leq n} be a finite sequence of nn real numbers >0> 0 such that ∑i=1nλi−1=1\sum_{i=1}^n \lambda_i^{-1} = 1. Show that r0≤sup⁡k(λk∣ak∣)1/kr_0 \leq \sup_k (\lambda_k |a_k|)^{1/k} [use a)].

c) Deduce from a) that, if the coefficients aia_i are all non-zero, we have

r0≤sup⁡(2∣a1∣,2∣a2a1∣,…,2∣an−1an−2∣,∣anan−1∣). r_0 \leq \sup \left( 2|a_1|, 2 \left| \frac{a_2}{a_1} \right|, \ldots, 2 \left| \frac{a_{n-1}}{a_{n-2}} \right|, \left| \frac{a_n}{a_{n-1}} \right| \right).

d) Deduce from a) that

r0≤∣a1−1∣+∣a2−a1∣+⋯+∣an−1−an∣+∣an∣ r_0 \leq |a_1 - 1| + |a_2 - a_1| + \cdots + |a_{n-1} - a_n| + |a_n|
[consider the polynomial (z−1)f(z)(z - 1)f(z)]. Hence show that, if the aia_i are all real and >0> 0, we have
r0≤sup⁡(a1,a2a1,…,an−1an−2,anan−1). r_0 \leq \sup \left( a_1, \frac{a_2}{a_1}, \ldots, \frac{a_{n-1}}{a_{n-2}}, \frac{a_n}{a_{n-1}} \right).

Solution

No solution has been written yet.

Transcribed from the printed volume. Not translated.